2002第17届CMO试题

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由式 ①，有
$P_{n+1}(x) P_{n-1}(x)-P_{n}^{2}(x)=
P_{n}(x) P_{n-2}(x)-P_{n-1}^{2}(x), n=3,4, \cdots$
$P_{n-1}(x)\left(P_{n+1}(x)+P_{n-1}(x)\right)=P_{n}(x)\left(P_{n}(x)+P_{n-2}(x)\right)$
由此递推可得
$\begin{aligned} \dfrac{P_{n+1}(x)}{P_{n}(x)}= \dfrac{P_{n}(x)+P_{n-2}(x)}{P_{n-1}(x)}=\cdots= \dfrac{P_{3}(x)+P_{1}(x)}{P_{2}(x)} \end{aligned}$ ②
因为 $P_{1}(x)=x^{2}-1, P_{2}(x)=2 x\left(x^{2}-1\right),P_{3}(x)=\dfrac{P_{2}^{2}(x)-\left(x^{2}-1\right)^{2}}{P_{1}(x)}=\left(x^{2}-1\right)\left(4 x^{2}-1\right)$
所以 $\dfrac{P_{3}(x)+P_{1}(x)}{P_{2}(x)}=\dfrac{\left(x^{2}-1\right)\left(4 x^{2}-1\right)+\left(x^{2}-1\right)}{2 x\left(x^{2}-1\right)}=2 x$ ③
将式 ③ 代人式 ②，得到 $P_{n+1}(x)=2 x P_{n}(x)-P_{n-1}(x), n=2,3, \cdots$ ④
补充定义 $P_0(x) = 0$．于是，式 ④ 对 $n = 1$ 也成立，式 ④ 对应的特征方程为 $\lambda^{2}-2 x \lambda+1=0$ 其根为 $\lambda=x \pm \sqrt{x^{2}-1}(|x|>1)$，易得 $P_{n}(x)=\dfrac{1}{2} \sqrt{x^{2}-1}\left(\left(x+\sqrt{x^{2}-1}\right)^{n}-\left(x-\sqrt{x^{2}-1}\right)^{2}\right)$ 其中 $n= 0,1,2, \cdots$ 由二项式定理，可得 $\displaystyle P_{n}(x)=\sum\limits_{m=0}^{\sigma_n} C_{n}^{2 m+1} x^{n-2 m-1}\left(x^{2}-1\right)^{m+1}$ 其中 $\sigma_{n}=\left\{\begin{array}{l}{\left[\frac{n}{2}\right], n为奇数} \\ {\frac{n}{2}-1, n为偶数}\end{array}\right.$ 则
$\displaystyle P_{n}(x)=\left(x^{2}-1\right) \sum\limits_{m=0}^{\sigma_{n}} C_{n}^{2 m+1} x^{n-2 m-1} \sum_{l=0}^{m}(-1)^{l} C_{m}^{l} x^{2 m-2 l}=\left(x^{2}-1\right) \sum_{l=0}^{\sigma_n}(-1)^{l} x^{n-2 l-1} \sum_{m=1}^{\sigma_n} C_{n}^{2 m+1} C_{m}^{l}$ ⑤
记 $\displaystyle a_{l}=\sum\limits_{m=l}^{\sigma_{\mathrm{n}}} \mathrm{C}_{n}^{2 m+1} \mathrm{C}_{m}^{l}$，代入式 ⑤ 得
$P_{n}(x)=\left(a_{0} x^{n+1}-a_{1} x^{n-1}+a_{2} x^{n-3}-\cdots+(-1)^{\sigma_n} a_{\sigma_n} x^{n-2\sigma_n +1}\right)-\\\left(a_{0} x^{n-1}-a_{1} x^{n-3}+\cdots+(-1)^{\sigma_{n}-1} a_{\sigma_{n}-1} x^{n-2 \sigma_n}+(-1)^{ \sigma_{n}} a_{\sigma_n} x^{n-2 \sigma_{n}-1} \right)$
由此可得
$\displaystyle S_{n}=a_{0}+\left(a_{1}+a_{0}\right)+\cdots+\left(a_{\sigma_{n}}+a_{\sigma_{n-1}}\right)+a_{\sigma_{n}}=2\left(a_{0}+a_{1}+\cdots+a_{\sigma_{n}}\right)\\=2 \sum\limits_{l=0}^{\sigma_n} \sum_{m=1}^{\sigma_{n}} C_{n}^{2 m+1} C_{m}^{l}=2 \sum\limits_{m=0}^{\sigma_{n}} C_{n}^{2 m+1} \sum_{l=0}^{\sigma_{m}} C_{m}^{l}=2 \sum\limits_{m=0}^{\sigma_{n}} C_{n}^{2 m+1} 2^{m}$
故 $S_{n}=\dfrac{1}{\sqrt{2}}\left[(1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right], n=0,1,2 \cdots$ ⑥
令 $t_{n}=(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n}, n=0,1,2, \cdots$，则 $S_{n}, t_{n}$ 都是非负整数，$S_{0}=0, S_{1}=2, t_{0}=t_{1}=2$，且 $t_{n}+\sqrt{2} S_{n}=2(1+\sqrt{2})^{n}, n=0,1,2, \cdots$ ⑦
当 $n \geqslant 2$ 时 $\displaystyle t_{n}=2 \sum\limits_{0 \leqslant 2 m \leqslant n} C_{n}^{2 m} 2^{n}=2\left(1+\sum_{0<2 m \leqslant n} C_{n}^{2 m} 2^{m}\right)$
可见，当 $n= 0,1,2, \cdots$ 时，$\dfrac{1}{2} t_{n}$ 为奇数
设 $n,m$ 为非负整数，由式 ⑦ 得
$\begin{aligned} t_{n+m}+\sqrt{2} S_{n+n}=& 2(1+\sqrt{2})^{n+m}= \dfrac{1}{2}\left(t_{n}+\sqrt{2} S_{n}\right)\left(t_{m}+\sqrt{2} S_{m}\right)= \dfrac{1}{2} t_{n} t_{m}+S_{n} S_{m}+\dfrac{\sqrt{2}}{2}\left(t_{n} S_{m}+t_{m} S_{n}\right) \end{aligned}$
从而，有 $t_{n+m}=\dfrac{1}{2} t_{n} t_{m}+S_{n} S_{m}, S_{n+m}=\dfrac{1}{2}\left(t_{n} S_{m}+t_{m} S_{n}\right)$ ⑧
令 $n= m$，得到 $S_{2 n}=\left(\dfrac{1}{2} t_{n}\right) 2 S_{n}, n=0,1,2, \cdots$
由于 $S_1 = 2$，而 $\dfrac{1}{2}t_n$ 为奇数，故由数学归纳法知 $2^{-(m+1)} S_{2}$ 为奇数，且 $m = 0,1,2, \cdots$，即
$k_{2^{m}}=m+1$ ⑨
再由式 ⑧ 可知，$k_n= k_m$，则 $k_{n+m}=\min \left\{k_{n}, k_{m}\right\}$ ⑩
对于任意正整数 $n$，设 $n=2^{m_{0}}+2^{m_{1}}+\cdots+2^{m_{l}}$
其中 $m_{0}, m_{1}, \cdots, m_{l}$ 均为整数，且 $0 \leqslant m_{0}<m_{1}<\cdots<m_{l}$．
由式 ⑨ 和式⑩可得，$k_{n}=m_{0}+1$．